Proof that $ frac{2^{-x}-1}{x} = sum_{n=0}^{infty} frac{ (-1)^{n+1}x^n(ln2)^{n+1}}{(n+1)!} $

Question

If
$$2^x = sum_{n=0}^{infty} frac{(xln(2))^n}{n!} hspace{1cm} forall x in mathbb{R},$$
Proof that:
$$ frac{2^{-x}-1}{x} =  sum_{n=0}^{infty} frac{ (-1)^{n+1}x^n(ln2)^{n+1}}{(n+1)!}  $$

I did the following:
begin{align*}
&2^x = sum_{n=0}^{infty} frac{(xln(2))^n}{n!} \
Rightarrow quad & 2^{-x} = frac{1}{sum_{n=0}^{infty} frac{(xln(2))^n}{n!}} \
Rightarrow quad & 2^{-x}-1 = frac{1}{sum_{n=0}^{infty} frac{(xln(2))^n}{n!}} -1 \
Rightarrow quad & dfrac{2^{-x}-1}{x}  = frac{frac{1}{sum_{n=0}^{infty} frac{(xln(2))^n}{n!}} -1}{x} \
Rightarrow quad & frac{2^{-x}-1}{x}  = frac{1-sum_{n=0}^{infty} frac{(xln(2))^n}{n!}}{xsum_{n=0}^{infty} frac{(xln(2))^n}{n!}}
end{align*}
From this part I don't know how to continue

Sebastiano · Answer

Your series can be written:
$$sum_{n=0}^{infty} frac{left(-1right)^{n + 1} x^{n} ln{left(2 right)}^{n + 1}}{left(n + 1right)!}=sum_{n=0}^{infty} frac{x^{n} left(- ln{left(2 right)}right)^{n + 1}}{left(n + 1right)!}$$
being $(-1)^{n+1}=-1$ forall $ninBbb N$.
And rewriting
$$sum_{n=0}^{infty} frac{x^{n} left(- ln{left(2 right)}right)^{n + 1}}{left(n + 1right)!}=sum_{n=0}^{infty} frac{left(- x ln{left(2 right)}right)^{n + 1}}{x left(n + 1right)!}$$
Shift the series by $1$ you obtain:
$$sum_{n=0}^{infty} frac{left(- x ln{left(2 right)}right)^{n + 1}}{x left(n + 1right)!}=sum_{n=1}^{infty} frac{left(- x ln{left(2 right)}right)^{n}}{x n!}=sum_{n=1}^{infty} frac{left(- x ln{left(2 right)}right)^{n}}{x n!}= frac{displaystylesum_{n=1}^{infty} frac{left(- x ln{left(2 right)}right)^{n}}{n!}}{x}$$
(remember that $x$ it is a constant).
After,
$$frac{sum_{n=1}^{infty} frac{left(- x ln{left(2 right)}right)^{n}}{n!}}x=frac{left(sum_{n=0}^{infty} frac{left(- x ln{left(2 right)}right)^{n}}{n!} + sum_{n=0}^{0} - frac{left(- x ln{left(2right)}right)^{n}}{n!}right)}{x}=frac{sum_{n=0}^{infty} frac{left(- x ln{left(2 right)}right)^{n}}{n!}-1}{x}=frac{e^{- x ln{left(2 right)}}-1}{x}$$
(you have an exponential series)
i.e.
$$frac{e^{- x ln{left(2 right)}}-1}{x}=frac{2^{-x}-1}{x}$$

Kirito · Answer

$$2^x = sum_{n=0}^{infty} dfrac{(xln(2))^n}{n!} $$
Proof:
$$2^{-x} = sum_{n=0}^{infty} dfrac{(-xln(2))^n}{n!} $$
$$Rightarrow 2^{-1x}  = 1 +sum_{n=1}^{infty} dfrac{(-1)^n(xln(2))^n}{n!} $$
$$Rightarrow 2^{-x} -1  = sum_{n=1}^{infty} dfrac{(-1)^n(xln(2))^n}{n!} $$
$$Rightarrow 2^{-x} -1  = sum_{n=0}^{infty} dfrac{(-1)^{n+1}(xln(2))^{n+1}}{(n+1)!} $$
$$Rightarrow  dfrac{2^{-x} -1}{x} = sum_{n=0}^{infty} dfrac{(-1)^{n+1}x^nln(2)^{n+1}}{(n+1)!} $$

user · Answer

We have
$$2^x = sum_{n=0}^{infty} dfrac{(xln(2))^n}{n!}$$
$$2^{-x} = sum_{n=0}^{infty} dfrac{(-x)^n(ln(2))^n}{n!}$$
$$2^{-x}-1 = -1+sum_{n=0}^{infty} dfrac{(-1)^nx^n(ln(2))^n}{n!}=sum_{n=0}^{infty} dfrac{(-1)^{n+1}x^{n+1}(ln(2))^{n+1}}{(n+1)!}$$
$$frac{2^{-x}-1}x =sum_{n=0}^{infty} dfrac{(-1)^{n+1}x^{n}(ln(2))^{n+1}}{(n+1)!}$$

Proof that $ frac{2^{-x}-1}{x} = sum_{n=0}^{infty} frac{ (-1)^{n+1}x^n(ln2)^{n+1}}{(n+1)!} $

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